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Question Number 168911 by bagjagugum123 last updated on 21/Apr/22

Commented by infinityaction last updated on 21/Apr/22

lim_(x→a) (((x^2 −a^2 )sin (x−a)((√(x+b))+(√(a+b)))^2 )/( ((√(x+b))−(√(a+b)))^2 ((√(x+b))+(√(a+b)))^2 )) lim_(x→a) (((x^2 −a^2 )sin (x−a)((√(x+b))+(√(a+b)))^2 )/((x−a)^2 )) lim_(x→a) (((x+a)sin (x−a)((√(x+b))+(√(a+b)))^2 )/((x−a))) 2a(2(√(a+b)))^2 8a(a+b)

$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\mathrm{sin}\:\left({x}−{a}\right)\left(\sqrt{{x}+{b}}+\sqrt{{a}+{b}}\right)^{\mathrm{2}} }{\:\left(\sqrt{{x}+{b}}−\sqrt{{a}+{b}}\right)^{\mathrm{2}} \left(\sqrt{{x}+{b}}+\sqrt{{a}+{b}}\right)^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\mathrm{sin}\:\left({x}−{a}\right)\left(\sqrt{{x}+{b}}+\sqrt{{a}+{b}}\right)^{\mathrm{2}} }{\left({x}−{a}\right)^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\left({x}+{a}\right)\mathrm{sin}\:\left({x}−{a}\right)\left(\sqrt{{x}+{b}}+\sqrt{{a}+{b}}\right)^{\mathrm{2}} }{\left({x}−{a}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}{a}\left(\mathrm{2}\sqrt{{a}+{b}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}{a}\left({a}+{b}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Answered by qaz last updated on 21/Apr/22

lim_(x→a) (((x^2 −a^2 )sin (x−a))/(((√(x+b))−(√(a+b)))^2 ))=lim_(x→0) (((x^2 +2ax)sin x)/(((√(x+a+b))−(√(a+b)))^2 )) =lim_(x→0) (((x^2 +2ax)sin x)/( (a+b)((√(1+(x/(a+b))))−1)^2 ))=lim_(x→0) ((2ax^2 +o(x^2 ))/((a+b)[(x^2 /(4(a+b)^2 ))+o(x^2 )])) =8a(a+b)

$$\underset{\mathrm{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\mathrm{sin}\:\left(\mathrm{x}−\mathrm{a}\right)}{\left(\sqrt{\mathrm{x}+\mathrm{b}}−\sqrt{\mathrm{a}+\mathrm{b}}\right)^{\mathrm{2}} }=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2ax}\right)\mathrm{sin}\:\mathrm{x}}{\left(\sqrt{\mathrm{x}+\mathrm{a}+\mathrm{b}}−\sqrt{\mathrm{a}+\mathrm{b}}\right)^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2ax}\right)\mathrm{sin}\:\mathrm{x}}{\:\left(\mathrm{a}+\mathrm{b}\right)\left(\sqrt{\mathrm{1}+\frac{\mathrm{x}}{\mathrm{a}+\mathrm{b}}}−\mathrm{1}\right)^{\mathrm{2}} }=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2ax}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{a}+\mathrm{b}\right)\left[\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} }+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)\right]} \\ $$$$=\mathrm{8a}\left(\mathrm{a}+\mathrm{b}\right) \\ $$

Commented by bagjagugum123 last updated on 21/Apr/22

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

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